2d Flashcards
organometallics havewhat type of preference
they have an e- count preference of either 18 ( Oh ) or 16 ( square planar )
normall organometallics want 18,, but what sometimes stop them from having 18e-
the metal!! early d block metals dont have enough e-
they normally need at least 6e- bc they get 12e- from th eligands
what happens to small metals like Ti and V when they form hexacarbonyl complexes and dont fill the 18e- preference
they get reduced to fulfill the 18e- rule
aka [Ti(CO)6]^2-
what happens to metals when they have a good e- count and acc form 18e- complexes with ligands (ligands provide 12e-)
they are very stable complexes bc all the 18e- go to the bonding orbitals and not no antibonding high energy eg* orbitals
what happens when the metals have too many e- and form complexes made up of 19e-
they dont do this
they dont mind less e- but they dont want more
bc theyre gonna go to the high energy eg* orbitals!! bc the 🔺o is so large
if Mn(CO)6 forms an 19e- complex,, can we form a Mn(CO)5 complex that will be 17e- ,, and then just reduce it to et 18e-
u could but we dont!!
we form a dimer.
aka we form a 2e- shared bond between the metals,, each metal will see the 2e- as theirs and therefore will count as 18e- for each metal.
so we trick the metals basically!!
so theres Oh and 18e- at each Mn.
wait can we acc get complexes with (cO)5 tho!!
yesss!!
if they have 18e- then theyre stable and are more than welcome to be a trigonal bipyramidal complex
okay so when we form a complex and theres 17e-,, what should we think
we should think of dimersiation
aka form a bond between the 2 metals. this will allow a neutral 18e- complex to form
whats a homoleptic complex
a complex where the metal is bonded to the same type of ligands
aka all the ligands are the same
what metals can form the dimer and which ones cant
Mn and Co can form dimers bc theyre large enough
but V cannot bc its too small and theres not enough space to form the dimer.
the metal being too small stops a dimer from being formed,, what else on a complex can cause the complex to not be able to form a dimer
if they have a super bulky and large ligand,, steric hinderance due to its size will prevent it from dimerising and forming a neutral 18e- complex.
can smt dimerise,, what do we think about
- does it have 17e-
- is the metal big enough
- are the ligands too bulky
how do we count e- in complexes
- we say every ligand donates 2e- each
- we look to see if the ligand is charged,, aka Br is Br-
- we look at the charged ligands and find the oxidation state of the metal by taking its overall charge into account.
- we find the metal oxidation state and remove this number from the metals e- count!!
- then we add the metals e- count and the ligands total e- count.
whats long u ,,, mu used for
it tells us if smt is bridging
long u2(CO) meaning
it means ur bridging 2 things with CO.
aka the CO is the bridge of choice
long u3(Me) meaning
ur bridging 3 things with an Me
aka the Me is the bridge of choice
whats the meaning of k,kappa
shows u the point of attachment!!!
nitrito - kN meaning
means the N of nitrito is bonded to the metal
nitrito-kO meaning
it means the O of the nitrito is bonded to the Metal
what does k1N(en) mean
it means the en ligand is bonded to the metal by the N ,, but its only bonded once
aka one bond from one N to the metal
what does k2N(en) mean
it means that in the ligand en,, 2 N are bonded to one metal.
so u have 2 bonds from en to the metal. 2N’s form a single bond each to the metal.
long n ( eta ) meaning
number of carbon atoms involvend in bonding!! but the carbon atoms are not independent,, theure seen as one.
aka like the zane salt and the
Pt, Cl3 bonded to the orthogonal ethene
in organometallic chemmm,, what is CO ,, aka a carbonyl
:C triple O
what bonds are seen in an organometallic carbonyl
1 sigma bond
2 pi bonds
tripl bond between the C and the O
